Question

prove that cosa/1+sina + 1+sina/cosa=1+sec​

Answer 1

Correct Questions:

To prove :

• $\frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } + \frac{1 + \sin( \alpha ) }{ \cos( \alpha ) } = 2 \sec( \alpha )$

Proof :

L.H.S. =

$= \frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } + \frac{1 + \sin( \alpha ) }{ \cos( \alpha ) }$

Taking L.C.M of Denominator and Solving,

We get,

$= \frac{ { \cos}^{2} \alpha + {(1 + \sin\alpha ) }^{2} }{ \cos \alpha (1 + \sin \alpha ) }$

But,

We know that,

• ${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$

Therefore,

We get,

$= \frac{ { \cos }^{2} \alpha + { \sin}^{2} \alpha + 2 \sin( \alpha ) + 1 } { \cos \alpha (1 + \sin\alpha ) }$

But,

We know that,

• ${ \sin }^{2} \beta + { \cos }^{2} \beta = 1$

Therefore,

We get,

$= \frac{1 +2 \sin( \alpha ) + 1}{ \cos\alpha (1 + \sin \alpha ) } \\ \\ = \frac{2 + 2\sin( \alpha ) }{ \cos \alpha (1 + \sin \alpha ) } \\ \\ = \frac{2(1 + \sin \alpha ) }{ \cos \alpha (1 + \sin \alpha ) } \\ \\ = \frac{2}{ \cos( \alpha ) }$

But,

We know that,

• $\frac{1}{ \cos( \beta ) } = \sec( \beta )$

Therefore,

We get,

$= 2 \sec( \alpha )$

= R.H.S

Since,

• L.H.S = R.H.S

Hence,

• Proved

Answer 2

Answer:

LHS=cosA/1+sinA+1+sinA/cosA

=cos^2A+1+2sinA+sin^2A/(1+sinA)cosA

=1+1+2sinA/(1+sinA)cosA

=2(1+sinA)/(1+sinA)cosA

1+1/cosA

1+secA=RHS=proved